SOLUTION :
Mass of 250 mL water = 250 g = 0.25 kg
Heat required to heat this water to 63ºC from 15ºC
= m s (t2 - t1)
= 0.25 * 4186 * (63 - 15)
= 50232 J
Let time taken is t sec by the 345 W immersion heater .
Electrical energy consumed
= 345 * t watt-sec
= 345 t J ( since 1 watt-sec = 1 J)
As per conservation of energy :
Electrical energy consumed = Heat energy generated to heat the water
=> 345 t = 50232
=> t = 50232 / 345 = 145.60 sec = 2 min. 25,60 secs.
Time needed to heat the water as required = 145.60 secs = 2 minutes 25.60 secs (ANSWER).
A small immersion heater is rated at 345 W. The specfic heat of water is 4186...
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